A More Accurate Half-Discrete Hilbert Inequality with a Nonhomogeneous Kernel

Abstract

By using the way of weight functions and the idea of introducing parameters, a more accurate half-discrete Hilbert inequality with a nonhomogeneous kernel and a best constant factor is presented. We also consider its best extension with the parameters, the equivalent forms, and the operator expressions as well as some reverses.

1. Introduction

If𝑎_𝑛,𝑏_𝑛≥0,  𝑎={𝑎_𝑛}^∞_𝑛=1∈𝑙²,  𝑏={𝑏_𝑛}^∞_𝑛=1∈𝑙²,  and  ‖𝑎‖={∑^∞_𝑛=1𝑎²_𝑛}^1/2>0  and  ‖𝑏‖={∑^∞_𝑛=1𝑏²_𝑛}^1/2>0,  then we have the following discrete Hilbert inequality (cf. [1]): ^∞∞_𝑛=1𝑚=1𝑎_𝑚𝑏_𝑛‖𝑎𝑏‖,𝑚+𝑛<𝜋‖‖ (1) where the constant factor𝜋is the best possible. Moreover, for𝑓, 𝑔(≥0)∈𝐿²(𝑅₊), ‖𝑓‖={∫^∞₀𝑓²(𝑥)𝑑𝑥}^1/2>0,  and  ‖𝑔‖>0,  we have the following Hilbert integral inequality (cf. [2]): ^∞₀𝑓(𝑥)𝑔𝑦𝑓𝑔,𝑥+𝑦𝑑𝑥𝑑𝑦<𝜋 (2) with the same best constant factor𝜋. Inequalities (1)-(2) are important in the analysis and its applications (cf. [3]). There are lots of improvements, generalizations, and applications of inequalities (1)-(2) for more details, refer to the literatures [4–15].

We find a result on the half-discrete Hilbert-type inequality with the nonhomogeneous kernel, which were published early as follows (cf. [16, Theorem 351]): ^∞₀𝑥^𝑞−2∞_𝑛=1𝑎_𝑛1+𝑥𝑛^𝑞𝑑𝑥<𝐵^𝑞1𝑝,1𝑞^∞_𝑛=1𝑎^𝑞_𝑛, (3) where𝑝>1,  𝑞>1,  1/𝑝+1/𝑞=1, and  𝐵(𝑢,𝑣)=∫^∞₀(𝑡^𝑢−1𝑑𝑡/(1+𝑡)^𝑢+𝑣)(𝑢,𝑣>0) is the beta function. But the constant factors 𝐵(1/𝑝,1/𝑞)were not considered whether the best possible. Recently, Yang gave a half-discrete Hilbert inequality with the nonhomogeneous kernel as follows (cf. [17]): ^∞_𝑛=1𝑎_𝑛^∞₀𝑓(𝑥)1+𝑥𝑛𝑑𝑥<𝜋^∞_𝑛=1𝑎²_𝑛^∞₀𝑓²(𝑥)𝑑𝑥^1/2, (4) where the constant factor 𝜋 is the best possible. And some half-discrete Hilbert-type inequalities with the nonhomogeneous kernel were given (cf. [18]).

In this paper, by using the way of weight functions and the idea of introducing parameters and by means of Hadamard’s inequality, we give the more accurate inequalities of (3) and (4) as follows: ^∞₀𝑥^𝑞−2∞_𝑛=1𝑎_𝑛1+𝑥𝑛−𝑥/2^𝑞𝑑𝑥<𝐵^𝑞1𝑝,1𝑞^∞_𝑛=1𝑎^𝑞_𝑛,_∞𝑛=1𝑎_𝑛^∞₀𝑓(𝑥)1+𝑥𝑛−𝑥/2𝑑𝑥<𝜋^∞_𝑛=1𝑎²_𝑛^∞₀𝑓²(𝑥)𝑑𝑥^1/2, (5)(6) with the best constant factors 𝐵(1/𝑝,1/𝑞) and 𝜋. We also consider their best extensions with the parameters, the equivalent forms, and the operator expressions as well as some reverses.

2. Some Lemmas

In the following lemmas, we assumed that𝛼, 𝛽, 𝜆>0,  𝛼<𝜆𝛽,  𝛼≤min{1,2−𝛽},  𝛾₁∈(−∞,∞), and  0≤𝛾₂≤1/2.

Lemma 1. Define the weight functions as follows: 𝜔_𝛼(𝑛)∶=𝑛−𝛾₂^𝛼∞_𝛾1𝑥−𝛾₁^𝛼−11+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆(),𝜛𝑑𝑥𝑛∈𝐍_𝛼(𝑥)∶=𝑥−𝛾₁^𝛼∞_𝑛=1𝑛−𝛾₂^𝛼−11+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆𝑥∈𝛾₁,∞. (7)(8) Setting 𝑘∶=𝑘(𝛼,𝛽,𝜆)=(1/𝛽)𝐵(𝛼/𝛽,𝜆−𝛼/𝛽), one has the following inequalities: 0<𝑘(1−𝜃(𝑥))<𝜛_𝛼(𝑥)<𝜔_𝛼(𝑛)=𝑘, (9) where 𝜃(𝑥)∶=(1/𝛽𝑘)∫^{(𝑥−𝛾1)𝛽(1−𝛾2)𝛽}₀(𝑢^{𝛼/𝛽−1}/(1+𝑢)^𝜆)𝑑𝑢∈(0,1)  and  𝜃(𝑥)=𝑂((𝑥−𝛾₁)^𝛼)  (𝑥∈(𝛾₁,∞)).

Proof. Putting 𝑢=(𝑥−𝛾₁)^𝛽(𝑛−𝛾₂)^𝛽 in (7), we have 𝜔_𝛼(𝑛)=1𝛽^∞₀𝑢^{𝛼/𝛽−1}(1+𝑢)^𝜆1𝑑𝑢=𝛽𝐵𝛼𝛽𝛼,𝜆−𝛽=𝑘. (10) For fixed 𝑥∈(𝛾₁,∞), setting 𝑓(𝑡)∶=𝑥−𝛾₁^𝛼𝑡−𝛾₂^𝛼−11+𝑥−𝛾₁^𝛽𝑡−𝛾₂^𝛽𝜆𝑡∈𝛾₂,∞, (11) in view of the conditions, we find that 𝑓^′(𝑡)=−(1−𝛼)(𝑥−𝛾₁)^𝛼(𝑡−𝛾₂)^𝛼−2/[1+(𝑥−𝛾₁)^𝛽(𝑡−𝛾₂)^𝛽]^𝜆−𝜆𝛽(𝑥−𝛾₁)^𝛼+𝛽(𝑡−𝛾₂)^{𝛼+𝛽−2}/[1+(𝑥−𝛾₁)^𝛽(𝑡−𝛾₂)^𝛽]^𝜆+1<0  and  𝑓^′′(𝑡)>0.  By the following Hadamard inequality (cf. [15]): 𝑓(𝑛)<^𝑛+1/2_𝑛−1/2𝑓(𝑡)𝑑𝑡(𝑛∈𝐍), (12) and letting  𝑢=(𝑥−𝛾₁)^𝛽(𝑡−𝛾₂)^𝛽,  it follows that 𝜛_𝛼(𝑥)=^∞_𝑛=1𝑓(𝑛)<^∞_𝑛=1^𝑛+1/2_𝑛−1/2𝑓(𝑡)𝑑𝑡=^∞_1/2𝑓(𝑡)≤𝑑𝑡_∞𝛾2𝑓(𝑡)1𝑑𝑡=𝛽^∞₀𝑢^{𝛼/𝛽−1}(1+𝑢)^𝜆=1𝑑𝑢𝛽𝐵𝛼𝛽𝛼,𝜆−𝛽𝜛=𝑘,_𝛼(𝑥)=^∞_𝑛=1𝑓(𝑛)>^∞₁𝑓(𝑡)=𝑑𝑡_∞𝛾2𝑓(𝑡)𝑑𝑡−¹_𝛾2𝑓(𝑡)1𝑑𝑡=𝑘−𝛽^{(𝑥−𝛾1)𝛽(1−𝛾2)𝛽}₀𝑢^{𝛼/𝛽−1}(1+𝑢)^𝜆((𝑥𝑑𝑢=𝑘1−𝜃))>0, (13) where (𝑥)=10<𝜃𝛽𝑘^{(𝑥−𝛾1)𝛽(1−𝛾2)𝛽}₀𝑢^{𝛼/𝛽−1}(1+𝑢)^𝜆<1𝑑𝑢𝛽𝑘^{(𝑥−𝛾1)𝛽(1−𝛾2)𝛽}₀𝑢^{𝛼/𝛽−1}𝑑𝑢=1−𝛾₂^𝛼𝑥−𝛾₁^𝛼.𝛼𝑘 (14) Hence, we prove that (9) is valid.

Lemma 2. Suppose that 1/𝑝+1/𝑞=1  (𝑝≠0,1), and 𝑎_𝑛≥0, and 𝑓(𝑥)≥0 is a real measurable function in (𝛾₁,∞). Then the following are considered. (i) For𝑝>1, one has the following: 𝐽∶=^∞_𝑛=1𝑛−𝛾₂^𝑝𝛼−1×^∞_𝛾1𝑓(𝑥)1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆𝑑𝑥^𝑝1/𝑝≤𝑘^1/𝑞∞_𝛾1𝜛_𝛼(𝑥)𝑥−𝛾₁^{𝑝(1−𝛼)−1}𝑓^𝑝(𝑥)𝑑𝑥^1/𝑝,𝐿₁∶=^∞_𝛾1𝑥−𝛾₁^𝑞𝛼−1𝜛_𝛼^𝑞−1(𝑥)×^∞_𝑛=1𝑎_𝑛1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆𝑞𝑑𝑥^1/𝑝≤𝑘^∞_𝑛=1𝑛−𝛾₂^{𝑞(1−𝛼)−1}𝑎^𝑞_𝑛^1/𝑞, (15)(16) where  𝜛_𝛼(𝑥) and 𝜔_𝛼(𝑛) are indicated by (7) and (8).
(ii) For𝑝<1  (𝑝≠0), one has the reverses of (15) and (16).

Proof. (i) By (7)–(9) and Hölder’s inequality (cf. [15]), we find that ^∞_𝛾1𝑓(𝑥)1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆𝑑𝑥^𝑝=^∞_𝛾111+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆×𝑥−𝛾₁^{(1−𝛼)/𝑞}𝑛−𝛾₂^{(1−𝛼)/𝑝}𝑓(𝑥×)𝑛−𝛾₂^{(1−𝛼)/𝑝}𝑥−𝛾₁^{(1−𝛼)/𝑞}𝑑𝑥^𝑝≤^∞_𝛾111+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆×𝑥−𝛾₁^{(1−𝛼)(𝑝−1)}𝑛−𝛾₂^1−𝛼𝑓^𝑝(𝑥)×𝑑𝑥_∞𝛾111+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆×𝑛−𝛾₂^{(1−𝛼)(𝑞−1)}𝑥−𝛾₁^1−𝛼𝑑𝑥^𝑝−1=^∞_𝛾1𝑓^𝑝(𝑥)𝑥−𝛾₁^{(1−𝛼)(𝑝−1)}1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆𝑑𝑥𝑛−𝛾₂^1−𝛼×𝑛−𝛾₂^{𝑞(1−𝛼)−1}𝜔_𝛼(𝑛)^𝑝−1=𝑛−𝛾₂^1−𝑝𝛼𝑘^𝑝−1×^∞_𝛾1𝑓^𝑝(𝑥)𝑥−𝛾₁^{(1−𝛼)(𝑝−1)}1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆1𝑛−𝛾₂^1−𝛼𝐽𝑑𝑥,_𝑝≤𝑘^∞𝑝−1_𝑛=1^∞_𝛾1𝑓^𝑝(𝑥)1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆×𝑥−𝛾₁^{(1−𝛼)(𝑝−1)}𝑛−𝛾₂^1−𝛼𝑑𝑥=𝑘^𝑝−1∞_𝛾1^∞_𝑛=1𝑛−𝛾₂^𝛼−11+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆×𝑥−𝛾₁^{𝛼+𝑝(1−𝛼)−1}𝑓^𝑝(𝑥)𝑑𝑥=𝑘^𝑝−1∞_𝛾1𝜛_𝛼(𝑥)𝑥−𝛾₁^{𝑝(1−𝛼)−1}𝑓^𝑝(𝑥)𝑑𝑥. (17)(18) Hence, (15) is valid. Using Hölder’s inequality again, we have ^∞_𝑛=1𝑎_𝑛1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆𝑞=^∞_𝑛=111+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆×𝑥−𝛾₁^{(1−𝛼)/𝑞}𝑛−𝛾₂^{(1−𝛼)/𝑝}𝑛−𝛾₂^{(1−𝛼)/𝑝}𝑥−𝛾₁^{(1−𝛼)/𝑞}𝑎_𝑛^𝑞≤𝜛_𝛼(𝑥)𝑥−𝛾₁^{𝑝(1−𝛼)−1𝑞−1}×^∞_𝑛=1𝑎^𝑞_𝑛1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆𝑛−𝛾₂^{(1−𝛼)𝑞/𝑝}𝑥−𝛾₁^1−𝛼=𝜛_𝛼^𝑞−1(𝑥)𝑥−𝛾₁^1−𝑞𝛼×^∞_𝑛=1𝑛−𝛾₂^{(1−𝛼)(𝑞−1)}1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆1𝑥−𝛾₁^1−𝛼𝑎^𝑞_𝑛,𝐿^𝑞₁≤^∞_𝛾1^∞_𝑛=111+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆×𝑛−𝛾₂^{(1−𝛼)(𝑞−1)}𝑥−𝛾₁^1−𝛼𝑎^𝑞_𝑛=𝑑𝑥_∞𝑛=1(𝑛−𝛾₂)^𝛼∞_𝛾1𝑥−𝛾₁^𝛼−11+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆×𝑑𝑥𝑛−𝛾₂^{𝑞(1−𝛼)−1}𝑎^𝑞_𝑛=^∞_𝑛=1𝜔_𝛼(𝑛)𝑛−𝛾₂^{𝑞(1−𝛼)−1}𝑎^𝑞_𝑛=𝑘^∞_𝑛=1𝑛−𝛾₂^{𝑞(1−𝛼)−1}𝑎^𝑞_𝑛. (19) Hence, (16) is valid.
(ii) For 0<𝑝<1  (𝑞<0) or 𝑝<0(0<𝑞<1), using the reverse Hölder inequality and in the same way, we have the reverses of (15) and (16).

3. Main Results

Some functions and spaces are introduced as follow: 𝜙(𝑥)∶=𝑥−𝛾₁^{𝑝(1−𝛼)−1},𝜙(𝑥)((𝑥𝜙(𝑥),𝜓(𝑛)∶=1−𝜃))∶=𝑛−𝛾₂^{𝑞(1−𝛼)−1}𝑥∈𝛾₁(𝑥),𝐿,∞,𝑛∈𝐍,𝜃isindicatedasLemma1_𝑝,𝜙𝛾₁,∞∶=𝑓;𝑓_𝑝,𝜙=^∞_𝛾1𝜙(𝑥)𝑓(𝑥)^𝑝𝑑𝑥^1/𝑝,𝑙<∞_𝑞,𝜓∶=𝑎=𝑎_𝑛^∞_𝑛=1;‖𝑎‖_𝑞,𝜓=^∞_𝑛=1𝜓(𝑛)𝑎_𝑛^𝑞1/𝑞<∞. (20)

Note 3. If 𝑝>1, then 𝐿_𝑝,𝜙(𝛾₁,∞) and 𝑙_𝑞,𝜓 are normal spaces; if 0<𝑝<1 or 𝑝<0, then both 𝐿_𝑝,𝜙(𝛾₁,∞) and 𝑙_𝑞,𝜓 are not normal spaces, but we still use the formal symbols in the following.

Theorem 3. Suppose that 𝑝>1,  1/𝑝+1/𝑞=1,  𝛼,𝛽,𝜆>0,  𝛼<𝜆𝛽,  𝛼≤min{1,2−𝛽},  𝛾₁∈(−∞,∞),  0≤𝛾₂≤1/2,  and 𝑓(𝑥),𝑎_𝑛≥0, such that𝑓∈𝐿_𝑝,𝜙(𝛾₁,∞),  𝑎={𝑎_𝑛}^∞_𝑛=1∈𝑙_𝑞,𝜓,  ‖𝑓‖_𝑝,𝜙>0,  and  ‖𝑎‖_𝑞,𝜓>0, then we have the following equivalent inequalities: 𝐼∶=^∞_𝑛=1𝑎_𝑛^∞_𝛾1𝑓(𝑥)1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆=𝑑𝑥_∞𝛾1𝑓(𝑥)^∞_𝑛=1𝑎_𝑛𝑑𝑥1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆<𝑘𝑓_𝑝,𝜙‖𝑎‖_𝑞,𝜓,𝐽=^∞_𝑛=1𝑛−𝛾₂^{𝑝𝛼−1∞}_𝛾1𝑓(𝑥)𝑑𝑥1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^{𝛽𝜆𝑝1/𝑝}<𝑘𝑓_𝑝,𝜙,𝐿=^∞_𝛾1(𝑥−𝛾₁)^{𝑞𝛼−1∞}_𝑛=1𝑎_𝑛1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆𝑞𝑑𝑥^1/𝑞<𝑘‖𝑎‖_𝑞,𝜓, (21)(22)(23) where the constant factor 𝑘=(1/𝛽)𝐵(𝛼/𝛽,𝜆−𝛼/𝛽) is the best possible.

Proof. By the Lebesgue term-by-term integration theorem, we find that there are two expressions of 𝐼 in (21). In view of (15), 𝜛_𝛼(𝑥)<𝑘, and 0<‖𝑓‖_𝑝,𝜙<∞, we have (22). By Hölder’s inequality, we find that 𝐼=^∞_𝑛=1𝑛−𝛾₂^{𝛼−1/𝑝∞}_𝛾1𝑓(𝑥)𝑑𝑥1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆×𝑛−𝛾₂^{1/𝑝−𝛼}𝑎_𝑛≤𝐽^∞_𝑛=1𝑛−𝛾₂^{𝑞(1−𝛼)−1}𝑎^𝑞_𝑛^1/𝑞=𝐽‖𝑎‖_𝑞,𝜓. (24) Hence, (21) is valid by (22). On the other hand, assuming that (21) is valid and setting 𝑎_𝑛∶=𝑛−𝛾₂^{𝑝𝛼−1∞}_𝛾1𝑓(𝑥)1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆𝑑𝑥^𝑝−1(𝑛∈𝐍), (25) then we have ‖𝑎‖^𝑞_𝑞,𝜓=^∞_𝑛=1𝑛−𝛾₂^{𝑞(1−𝛼)−1}𝑎^𝑞_𝑛=𝐽^𝑝=𝐼. (26) By (15), it follows that 𝐽<∞. If 𝐽=0, then (22) is trivially valid. If 𝐽>0, then 0<‖𝑎‖_𝑞,𝜓=𝐽^𝑝−1<∞. By (21), we have ‖𝑎‖^𝑞_𝑞,𝜓=𝐽^𝑝=𝐼<𝑘𝑓_𝑝,𝜙‖𝑎‖_𝑞,𝜓,thatis,𝐽=‖𝑎‖^𝑞−1_𝑞,𝜓<𝑘𝑓_𝑝,𝜙. (27) Hence, (22) is valid, which is equivalent to (21).
In view of (16) and 𝜛_𝛼(𝑥)<𝑘, we obtain (23). By Hölder’s inequality again, we have 𝐼=^∞_𝛾1𝑥−𝛾₁^{∞𝛼−1/𝑞}_𝑛=1𝑎_𝑛1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆×𝑥−𝛾₁^{1/𝑞−𝛼}𝑓(𝑥)𝑑𝑥≤𝐿^∞_𝛾1𝑥−𝛾₁^{𝑝(1−𝛼)−1}𝑓^𝑝(𝑥)𝑑𝑥^1/𝑝=𝐿𝑓_𝑝,𝜙. (28) Then (21) is valid by using (23). Supposing that (21) is valid and setting 𝑓(𝑥)∶=𝑥−𝛾₁^{𝑞𝛼−1∞}_𝑛=1𝑎_𝑛1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^{𝛽𝜆𝑞−1}𝑥∈𝛾₁,∞, (29) then it follows that 𝑓^𝑝_𝑝,𝜙=^∞_𝛾1𝑥−𝛾₁^{𝑝(1−𝛼)−1}𝑓^𝑝(𝑥)𝑑𝑥=𝐿^𝑞=𝐼. (30) In view of (16) and (9), we find 𝐿<∞. If 𝐿=0, then (23) is trivially valid; if 𝐿>0, that is,0<‖𝑓‖_𝑝,𝜙<∞, then, by (21), we have 𝑓^𝑝_𝑝,𝜙=𝐿^𝑞=𝐼<𝑘𝑓_𝑝,𝜙‖𝑎‖_𝑞,𝜓,thatis,𝐿=𝑓^𝑝−1_𝑝,𝜙<𝑘‖𝑎‖_𝑞,𝜓. (31) Hence (23) is valid, which is equivalent to (21). Then (21), (22), and (23) are equivalent. We prove that the constant factor in (21) is the best possible. For 0<𝜀<𝑞𝛼, setting 𝑎={𝑎_𝑛}^∞_𝑛=1 and 𝑓(𝑥) as follows (𝑝,𝑞>1): 𝑎_𝑛=𝑛−𝛾₂^{𝛼−𝜀/𝑞−1},𝑓(𝑥)=𝑥−𝛾₁^{𝛼+𝜀/𝑝−1},𝑥∈𝛾₁,1+𝛾₁,0,𝑥∈1+𝛾₁,∞, (32) if there exists a positive number 𝑘^′≤𝑘, such that (21) is still valid as we replace 𝑘 by 𝑘^′, then, substitution of 𝑎 and 𝑓(𝑥), we have 𝐼̃∶=^∞_𝑛=1𝑎_𝑛^∞_𝛾1𝑓(𝑥)1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆𝑑𝑥<𝑘^′𝑓_𝑝,𝜙‖𝑎‖_𝑞,𝜓. (33)
For 𝑝, 𝑞>1, letting 𝛼=𝛼−𝜀/𝑞, we find by Lemma 1 that 𝐼̃=^𝛾1_𝛾⁺¹₁𝑥−𝛾₁^𝜀−1𝑥−𝛾₁^{𝛼−𝜀/𝑞}×^∞_𝑛=1𝑛−𝛾₂^{𝛼−𝜀/𝑞−1}𝑑𝑥1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆=^𝛾1_𝛾⁺¹₁𝑥−𝛾₁^𝜀−1𝜛𝛼(𝑥)𝑑𝑥>𝑘𝛼,𝛽,𝜆^𝛾1_𝛾⁺¹₁𝑥−𝛾₁^𝜀−11−𝑂𝑥−𝛾₁𝛼𝜀𝑑𝑥=𝑘𝛼−𝑞1,𝛽,𝜆𝜀−𝑂(1). (34) By simple calculation, we have ‖𝑎‖^𝑞_𝑞,𝜓=^∞_𝑛=1𝑛−𝛾₂^{𝑞(1−𝛼)−1}𝑎^𝑞_𝑛=^∞_𝑛=1𝑛−𝛾₂^−1−𝜀<1−𝛾₂^−1−𝜀+^∞₁𝑦−𝛾₂^−1−𝜀𝑑𝑦=𝜀+1−𝛾₂𝜀1−𝛾₂^𝜀+1,𝑓_𝑝,𝜙=^𝛾1_𝛾⁺¹₁𝑥−𝛾₁^𝜀−1𝑑𝑥^1/𝑝=1𝜀^1/𝑝. (35)(36) In view of (33), (34), and 𝑞>1, it follows that 𝑘𝜀𝛼−𝑞[(1̃,𝛽,𝜆1−𝜀𝑂)]<𝜀𝐼<𝑘^′𝑓_𝑝,𝜙‖𝑎‖_𝑞,𝜓<𝑘^′𝜀+1−𝛾₂1−𝛾₂^𝜀+11/𝑞. (37) Then we have 𝑘(𝛼,𝛽,𝜆)=𝑘≤𝑘^′ for 𝜀→0⁺. Hence, 𝑘^′=𝑘 is the best value of (21). By the equivalency, the constant factor 𝑘 in (22) ((23)) is the best possible. Otherwise, we would reach a contradiction by (24) ((28)) that the constant factor in (21) is not the best possible.

Remark 4. (i) Define a half-discrete Hilbert operator 𝑇∶𝐿_𝑝,𝜙(𝛾₁,∞)→𝑙_{𝑝,𝜓1−𝑝} as follows. For 𝑓∈𝐿_𝑝,𝜙(𝛾₁,∞), we define 𝑇𝑓∈𝑙_{𝑝,𝜓1−𝑝}, satisfying 𝑇𝑓(𝑛)=^∞_𝛾1𝑓(𝑥)1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆𝑑𝑥(𝑛∈𝐍). (38) Then by (22), it follows that ‖𝑇𝑓‖_{𝑝,𝜓1−𝑝}≤𝑘‖𝑓‖_𝑝,𝜙; that is, 𝑇 is the bounded operator with ‖𝑇‖≤𝑘. Since the constant factor𝑘in (22) is the best possible, then we have ‖𝑇‖=𝑘=(1/𝛽)𝐵(𝛼/𝛽,𝜆−𝛼/𝛽).
(ii) Define a half-discrete Hilbert operator 𝑇∶𝑙_𝑞,𝜓→𝐿_𝑞,𝜙^1−𝑞(𝛾₁,∞) in the following way. For 𝑎∈𝑙_𝑞,𝜓, we define 𝑇𝑎∈𝐿_𝑞,𝜙^1−𝑞, satisfying 𝑇𝑎(𝑥)=^∞_𝑛=1𝑎_𝑛1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆𝑥∈𝛾₁,∞. (39) Then by (23), it follows that ‖𝑇𝑎‖_𝑞,𝜙^1−𝑞≤𝑘‖𝑎‖_𝑞,𝜓; that is, 𝑇 is the bounded operator with ‖𝑇‖≤𝑘. Since the constant factor 𝑘 in (23) is the best possible, then we have ‖𝑇‖=𝑘=(1/𝛽)𝐵(𝛼/𝛽,𝜆−𝛼/𝛽).

Theorem 5. Suppose that 0<𝑝<1, 1/𝑝+1/𝑞=1, 𝛼,𝛽,𝜆>0, 𝛼<𝜆𝛽, 𝛼≤min{1,2−𝛽}, 𝛾₁∈(−∞,∞), 0≤𝛾₂≤1/2, and  𝑓(𝑥),𝑎_𝑛≥0, such that 𝑓∈𝐿_𝑝,𝜙(𝛾₁,∞), 𝑎={𝑎_𝑛}^∞_𝑛=1∈𝑙_𝑞,𝜓, ‖𝑓‖_𝑝,𝜙>0, and ‖𝑎‖_𝑞,𝜓>0. Then one has the following equivalent inequalities: 𝐼=^∞_𝑛=1𝑎_𝑛^∞_𝛾1𝑓(𝑥)1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆=𝑑𝑥_∞𝛾1𝑓(𝑥)^∞_𝑛=1𝑎_𝑛𝑑𝑥1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆>𝑘𝑓_𝑝,𝜙‖𝑎‖_𝑞,𝜓,𝐽=^∞_𝑛=1𝑛−𝛾₂^{𝑝𝛼−1∞}_𝛾1𝑓(𝑥)1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆𝑑𝑥^𝑝1/𝑝>𝑘𝑓_𝑝,𝜙,𝐿∶=^∞_𝛾1𝑥−𝛾₁^𝑞𝛼−1[1−𝜃(𝑥)]^𝑞−1∞_𝑛=1𝑎_𝑛[1+(𝑥−𝛾₁)^𝛽(𝑛−𝛾₂)^𝛽]^𝜆𝑞𝑑𝑥^1/𝑞>𝑘‖𝑎‖_𝑞,𝜓, (40)(41)(42) where the constant factor 𝑘=(1/𝛽)𝐵(𝛼/𝛽,𝜆−𝛼/𝛽) is the best possible.

Proof. In view of (9), the reverse of (15), and 0<‖𝑓‖_𝑝,𝜙<∞, we have (41). Using the reverse Hölder inequality, we obtain the reverse form of (24) as follows: 𝐼≥𝐽‖𝑎‖_𝑞,𝜓. (43) Then by (41), (40) is valid.
On the other hand, if (40) is valid, setting 𝑎_𝑛 as in (25), then (26) still holds with 0<𝑝<1. By (40), it follows that 𝐽>0. If 𝐽=∞, then (41) is trivially valid; if 𝐽<∞, then 0<‖𝑎‖_𝑞,𝜓=𝐽^𝑝−1<∞, and we have ‖𝑎‖^𝑞_𝑞,𝜓=𝐽^𝑝=𝐼>𝑘𝑓_𝑝,𝜙‖𝑎‖_𝑞,𝜓,thatis,𝐽=‖𝑎‖^𝑞−1_𝑞,𝜓>𝑘𝑓_𝑝,𝜙. (44) Then (41) is valid, which is equivalent to (40).
By the reverse of (16) and 𝜛(𝑥)>𝑘(1−𝜃(𝑥)), and by 𝑞<0, we have 𝐿>𝑘^{(𝑞−1)/𝑞}𝐿₁≥𝑘^{(𝑞−1)/𝑞}𝑘^∞_𝑛=1𝑛−𝛾₂^{𝑞(1−𝛼)−1}𝑎^𝑞_𝑛^1/𝑞=𝑘‖𝑎‖_𝑞,𝜓. (45) Hence, (42) is valid. By the reverse Hölder inequality again, we obtain 𝐼=^∞_𝛾1𝑥−𝛾₁^{𝛼−1/𝑞}(1−𝜃(𝑥))^∞1/𝑝_𝑛=1𝑎_𝑛1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆×(1−𝜃(𝑥))^1/𝑝𝑥−𝛾₁^{1/𝑞−𝛼}𝑓(𝑥≥𝐿𝑓)𝑑𝑥_𝑝,𝜙. (46) Then (40) is valid by (42). On the other hand, if (40) is valid, setting 𝑓(𝑥)=𝑥−𝛾₁^𝑞𝛼−1[1−𝜃(𝑥)]^𝑞−1∞_𝑛=1𝑎_𝑛1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^{𝛽𝜆𝑞−1}𝑥∈𝛾₁,∞, (47) then ‖𝑓‖^𝑝_𝑝,𝜙=∫^∞_𝛾1[1−𝜃(𝑥)](𝑥−𝛾₁)^{𝑝(1−𝛼)−1}𝑓^𝑝(𝑥)𝑑𝑥=𝐿^𝑞=𝐼. By the reverse of (16), we have 𝐿>0. If 𝐿=∞, then (42) is trivially valid; if 0<𝐿<∞, then, by (40), we obtain 𝑓^𝑝_𝑝,𝜙=𝐿^𝑞=𝐼>𝑘𝑓_𝑝,𝜙‖𝑎‖_𝑞,𝜓,thatis,𝐿=𝑓^𝑝−1_𝑝,𝜙>𝑘‖𝑎‖_𝑞,𝜓. (48) Hence, (42) is valid, which is equivalent to (40). It follows that (40), (41), and (42) are equivalent.
If there exists a positive number 𝑘^′≥𝑘, such that (40) is still valid as we replace 𝑘 by 𝑘^′, then, in particular, we have 𝐼̃=^∞_𝑛=1𝑎_𝑛^∞_𝛾1𝑓(𝑥)1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆𝑑𝑥>𝑘^′𝑓_𝑝,𝜙‖𝑎‖_𝑞,𝜓, (49) where 𝑎={𝑎_𝑛}^∞_𝑛=1 and 𝑓 are taken as in (32) (0<𝜀<𝑝(𝛽𝜆−𝛼)). Since, by (35), we find in 𝑓_𝑝,𝜙=^1+𝛾1_𝛾11−𝑂𝑥−𝛾₁^𝛼𝑥−𝛾₁^𝜀−1𝑑𝑥^1/𝑝=1𝜀−𝑂(1)^1/𝑝,𝐼̃≤^∞_𝑛=1𝑛−𝛾₂^{𝛼−𝜀/𝑞−1∞}_𝛾1𝑥−𝛾₁^{𝛼+𝜀/𝑝−1}1+𝑥−𝛾₁^𝛽𝑛−𝛾₂^𝛽𝜆=1𝑑𝑥𝛽^∞_𝑛=1𝑛−𝛾₂^{−𝜀−1∞}₀𝑢^{𝛼/𝛽+𝜀/𝛽𝑝−1}(1+𝑢)^𝜆≤1𝑑𝑢𝛽𝜀+1−𝛾₂𝜀1−𝛾₂^𝜀+1𝐵𝛼𝛽+𝜀𝛼𝛽𝑝,𝜆−𝛽−𝜀,𝛽𝑝 (50) then, by (35), (49), and the above results, we obtain (notice that𝑞<0) 1𝛽𝜀+1−𝛾₂1−𝛾₂^𝜀+1𝐵𝛼𝛽+𝜀𝛼𝛽𝑝,𝜆−𝛽−𝜀̃𝛽𝑝≥𝜀𝐼>𝜀𝑘^′1𝜀−𝑂(1)^1/𝑝𝜀+1−𝛾₂𝜀1−𝛾₂^𝜀+11/𝑞=𝑘^′(1−𝜀𝑂(1))^1/𝑝𝜀+1−𝛾₂1−𝛾₂^𝜀+11/𝑞. (51) For 𝜀→0⁺, we have 𝑘=(1/𝛽)𝐵(𝛼/𝛽,𝜆−𝛼/𝛽)≥𝑘^′. Hence, 𝑘=𝑘^′ is the best value of (40). By the equivalency, the constant factor 𝑘 in (41) ((42)) is the best possible. Otherwise, we would reach a contradiction by (43) ((46)) that the constant factor in (40) is not the best possible.

In the same way, for 𝑝<0, we also have the following result.

Theorem 6. If the assumption of 𝑝>1 in Theorem 3 is replaced by 𝑝<0, then the reverses of (21), (22), and (23) are valid and equivalent. Moreover, the same constant factor is the best possible.

Remark 7. (i) If 𝛾₁=𝛾₂=0,  𝜆=𝛽=1, and  𝛼=1/𝑝, then𝑘=𝐵(1/𝑝,1/𝑞) and (23) reduces to (3). In particular, for 𝑝=𝑞=2, (21) reduces to (4).
(ii) For𝛾₁=0,  𝛾₂=1/2, and  𝛼=1/𝑝in (21) and (23), it follows ^∞_𝑛=1𝑎_𝑛^∞₀𝑓(𝑥)1+(𝑥𝑛−𝑥/2)^𝛽𝜆𝑑𝑥<𝑘^∞₀𝑥^𝑝−2𝑓^𝑝(𝑥)𝑑𝑥^1/𝑝∞_𝑛=1𝑎^𝑞_𝑛^1/𝑞,_∞0𝑥^𝑞−2∞_𝑛=1𝑎_𝑛1+(𝑥𝑛−𝑥/2)^𝛽𝜆𝑞𝑑𝑥<𝑘^𝑞∞_𝑛=1𝑎^𝑞_𝑛, (52)(53) where the constant factors both 𝑘=(1/𝛽)𝐵(1/𝑝𝛽,𝜆−1/𝑝𝛽) and 𝑘^𝑞 are the best possibles. In particular, for 𝜆=𝛽=1, (53) comes to (5); for 𝜆=𝛽=1 and  𝑝=𝑞=2, in (52), we obtain (6). Hence, inequality (21) is the best extension of (4) and (6), and inequality (23) is the best extension of (3) and (5) with the parameters.

Acknowledgment

This paper is supported by the 2012 Knowledge Construction Special Foundation Item of Guangdong Institution of Higher Learning College and University (no. 2012KJCX0079).